Properties of Inverse Trigonometric Functions — Worked Solutions

Ques­tion 1: Four inverse tan­gents add to $\displaystyle \frac{\pi}{4}$#

The prob­lem#

Prove: $\displaystyle \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{7} + \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{8} = \frac{\pi}{4}$.

Under­stand­ing the prob­lem#

We must com­bine four inverse tan­gents into one and show it equals $\displaystyle \tfrac{\pi}{4} = \tan^{-1}1$.

The idea#

Use $\displaystyle \tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x + y}{1 - xy}$ (valid when $xy < 1$) twice in pairs, and then once more.

Step-by-step solu­tion#

Step 1. Com­bine the first pair. Here $\displaystyle xy = \tfrac{1}{35} < 1$.

$$\displaystyle \tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{7} = \tan^{-1}\frac{\frac{1}{5} + \frac{1}{7}}{1 - \frac{1}{35}} = \tan^{-1}\frac{\frac{12}{35}}{\frac{34}{35}} = \tan^{-1}\frac{12}{34} = \tan^{-1}\frac{6}{17}$$

Step 2. Com­bine the sec­ond pair. Here $\displaystyle xy = \tfrac{1}{24} < 1$.

$$\displaystyle \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{8} = \tan^{-1}\frac{\frac{11}{24}}{\frac{23}{24}} = \tan^{-1}\frac{11}{23}$$

Step 3. Com­bine the two results. Here $\displaystyle xy = \tfrac{66}{391} < 1$.

$$\displaystyle \tan^{-1}\frac{6}{17} + \tan^{-1}\frac{11}{23} = \tan^{-1}\frac{\frac{6}{17} + \frac{11}{23}}{1 - \frac{66}{391}}$$

Step 4. Mul­ti­ply top and bot­tom by $17 \times 23 = 391$.

$$\displaystyle = \tan^{-1}\frac{6 \cdot 23 + 11 \cdot 17}{391 - 66} = \tan^{-1}\frac{138 + 187}{325} = \tan^{-1}\frac{325}{325} = \tan^{-1}1 = \frac{\pi}{4}$$

Check­ing the answer#

Approx­i­mate val­ues: $\displaystyle 0.1974 + 0.1419 + 0.3218 + 0.1244 = 0.7855 \approx \tfrac{\pi}{4} = 0.7854$.

Answer#

The sum is $\displaystyle \tan^{-1}1 = \frac{\pi}{4}$, as required.

Ques­tion 2: A com­bi­na­tion of prin­ci­pal val­ues#

The prob­lem#

Find $\displaystyle \cos^{-1}\left(-\frac{1}{2}\right) - 2\sin^{-1}\frac{1}{2}$.

Under­stand­ing the prob­lem#

We need the prin­ci­pal val­ues: $\cos^{-1}$ in $[0, \pi]$ and $\sin^{-1}$ in $\displaystyle \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$.

The idea#

Use $\cos^{-1}(-x) = \pi - \cos^{-1}x$ for the neg­a­tive argu­ment.

Step-by-step solu­tion#

Step 1. $\displaystyle \cos^{-1}\frac{1}{2} = \frac{\pi}{3}$, so

$$\displaystyle \cos^{-1}\left(-\frac{1}{2}\right) = \pi - \frac{\pi}{3} = \frac{2\pi}{3}$$

Step 2. $\displaystyle \sin^{-1}\frac{1}{2} = \frac{\pi}{6}$, so $\displaystyle 2\sin^{-1}\frac{1}{2} = \frac{\pi}{3}$.

Step 3. Sub­tract.

$$\displaystyle \frac{2\pi}{3} - \frac{\pi}{3} = \frac{\pi}{3}$$

Check­ing the answer#

$\displaystyle \cos\frac{2\pi}{3} = -\frac{1}{2}$ and $\displaystyle \frac{2\pi}{3}$ lies in $[0, \pi]$, so Step 1 is the prin­ci­pal value.

Answer#

$\displaystyle \frac{\pi}{3}$

Com­mon mis­take to avoid#

Writ­ing $\displaystyle \cos^{-1}\left(-\tfrac{1}{2}\right) = -\tfrac{\pi}{3}$. That is out­side the range $[0, \pi]$.

Ques­tion 3: Sim­pli­fy­ing with half angles#

The prob­lem#

Sim­plify $\displaystyle \tan^{-1}\frac{\cos x}{1 - \sin x}$ for $\displaystyle -\frac{\pi}{2} < x < \frac{\pi}{2}$.

Under­stand­ing the prob­lem#

We want to write the frac­tion as $\tan(\text{something})$, so the $\tan^{-1}$ and $\tan$ can­cel. We must then check that "some­thing" lies in $\displaystyle \left(-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right)$.

The idea#

Write $\cos x$ and $1 - \sin x$ with half angles, using $\displaystyle c = \cos\tfrac{x}{2}$ and $\displaystyle s = \sin\tfrac{x}{2}$:
$\cos x = c^2 - s^2$ and $1 - \sin x = c^2 + s^2 - 2sc = (c - s)^2$.

Step-by-step solu­tion#

Step 1. Rewrite the frac­tion.

$$\displaystyle \frac{\cos x}{1 - \sin x} = \frac{(c - s)(c + s)}{(c - s)^2} = \frac{c + s}{c - s}$$

Step 2. Divide top and bot­tom by $\displaystyle c = \cos\tfrac{x}{2}$ (not zero in this range).

$$\displaystyle = \frac{1 + \tan\frac{x}{2}}{1 - \tan\frac{x}{2}} = \tan\left(\frac{\pi}{4} + \frac{x}{2}\right)$$

(This is $\tan(A + B)$ with $\displaystyle \tan A = \tan\tfrac{\pi}{4} = 1$.)

Step 3. Check the range. From $\displaystyle -\tfrac{\pi}{2} < x < \tfrac{\pi}{2}$ we get $\displaystyle 0 < \tfrac{\pi}{4} + \tfrac{x}{2} < \tfrac{\pi}{2}$, which is inside the prin­ci­pal range of $\tan^{-1}$.

$$\displaystyle \tan^{-1}\frac{\cos x}{1 - \sin x} = \frac{\pi}{4} + \frac{x}{2}$$

Check­ing the answer#

At $x = 0$: $\displaystyle \tan^{-1}\frac{1}{1} = \frac{\pi}{4}$, and $\displaystyle \frac{\pi}{4} + 0 = \frac{\pi}{4}$.

Answer#

$\displaystyle \frac{\pi}{4} + \frac{x}{2}$

Ques­tion 4: Sim­pli­fy­ing by sub­sti­tu­tion#

The prob­lem#

Sim­plify $\displaystyle \tan^{-1}\frac{3a^2x - x^3}{a^3 - 3ax^2}$ for $\displaystyle \lvert x \rvert < \frac{a}{\sqrt{3}}$, $a > 0$.

Under­stand­ing the prob­lem#

The expres­sion looks like the for­mula $\displaystyle \tan 3\theta = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}$. The given range for $x$ will make sure the answer lies in the prin­ci­pal range.

The idea#

Sub­sti­tute $x = a\tan\theta$, so $\displaystyle \theta = \tan^{-1}\frac{x}{a}$.

Step-by-step solu­tion#

Step 1. Sub­sti­tute $x = a\tan\theta$.

$$\displaystyle \frac{3a^2 \cdot a\tan\theta - a^3\tan^3\theta}{a^3 - 3a \cdot a^2\tan^2\theta} = \frac{a^3(3\tan\theta - \tan^3\theta)}{a^3(1 - 3\tan^2\theta)} = \tan 3\theta$$

Step 2. Check the range. $\displaystyle \lvert x \rvert < \frac{a}{\sqrt{3}}$ means $\displaystyle \lvert\tan\theta\rvert < \frac{1}{\sqrt{3}}$, so $\displaystyle -\frac{\pi}{6} < \theta < \frac{\pi}{6}$ and $\displaystyle -\frac{\pi}{2} < 3\theta < \frac{\pi}{2}$.

Step 3. So $\tan^{-1}(\tan 3\theta) = 3\theta$.

$$\displaystyle \tan^{-1}\frac{3a^2x - x^3}{a^3 - 3ax^2} = 3\theta = 3\tan^{-1}\frac{x}{a}$$

Check­ing the answer#

At $x = 0$ both sides are $0$.

Answer#

$\displaystyle 3\tan^{-1}\frac{x}{a}$

Ques­tion 5: Trigono­met­ric val­ues of inverse expres­sions#

The prob­lem#

Find (a) $\displaystyle \sin\left(2\tan^{-1}\frac{1}{3}\right)$ and (b) $\displaystyle \cos\left(\sin^{-1}\frac{8}{17} + \cos^{-1}\frac{3}{5}\right)$.

Under­stand­ing the prob­lem#

Name the inner angles and use dou­ble-angle or sum for­mu­las. All the angles here are acute because the argu­ments are pos­i­tive.

The idea#

(a) Let $\displaystyle \theta = \tan^{-1}\tfrac{1}{3}$ and use $\displaystyle \sin 2\theta = \frac{2\tan\theta}{1 + \tan^2\theta}$.
(b) Let $\displaystyle A = \sin^{-1}\tfrac{8}{17}$, $\displaystyle B = \cos^{-1}\tfrac{3}{5}$ and use $\cos(A + B) = \cos A\cos B - \sin A\sin B$.

Step-by-step solu­tion#

Part (a)

Step 1. With $\displaystyle \tan\theta = \tfrac{1}{3}$:

$$\displaystyle \sin 2\theta = \frac{2 \cdot \frac{1}{3}}{1 + \frac{1}{9}} = \frac{\frac{2}{3}}{\frac{10}{9}} = \frac{2}{3}\cdot\frac{9}{10} = \frac{3}{5}$$

Part (b)

Step 1. $\displaystyle \sin A = \tfrac{8}{17}$, so $\displaystyle \cos A = \sqrt{1 - \tfrac{64}{289}} = \tfrac{15}{17}$ (an $8$-$15$-$17$ tri­an­gle).

Step 2. $\displaystyle \cos B = \tfrac{3}{5}$, so $\displaystyle \sin B = \tfrac{4}{5}$ (a $3$-$4$-$5$ tri­an­gle).

Step 3. Apply the for­mula.

$$\displaystyle \cos(A + B) = \frac{15}{17}\cdot\frac{3}{5} - \frac{8}{17}\cdot\frac{4}{5} = \frac{45}{85} - \frac{32}{85} = \frac{13}{85}$$

Check­ing the answer#

(a) $\displaystyle \tan^{-1}\tfrac{1}{3} \approx 18.43^\circ$; $\displaystyle \sin 36.87^\circ \approx 0.6 = \tfrac{3}{5}$.

Answer#

(a) $\displaystyle \frac{3}{5}$ (b) $\displaystyle \frac{13}{85}$

Ques­tion 6: An equa­tion in inverse tan­gents#

The prob­lem#

Solve $\displaystyle \tan^{-1}\frac{x - 1}{x - 2} + \tan^{-1}\frac{x + 1}{x + 2} = \frac{\pi}{4}$.

Under­stand­ing the prob­lem#

Com­bine the two inverse tan­gents into one, then take $\tan$ of both sides. At the end, check each root in the orig­i­nal equa­tion.

The idea#

$\displaystyle \tan^{-1}p + \tan^{-1}q = \tan^{-1}\frac{p + q}{1 - pq}$, and $\displaystyle \tan\tfrac{\pi}{4} = 1$.

Step-by-step solu­tion#

Step 1. With $\displaystyle p = \frac{x - 1}{x - 2}$, $\displaystyle q = \frac{x + 1}{x + 2}$, the equa­tion becomes $\displaystyle \frac{p + q}{1 - pq} = 1$.

Step 2. Mul­ti­ply top and bot­tom by $(x - 2)(x + 2)$.

$$\displaystyle \frac{(x - 1)(x + 2) + (x + 1)(x - 2)}{(x - 2)(x + 2) - (x - 1)(x + 1)} = 1$$

Step 3. Expand.

$$\begin{aligned} \text{top} &= (x^2 + x - 2) + (x^2 - x - 2) = 2x^2 - 4\\ \text{bottom} &= (x^2 - 4) - (x^2 - 1) = -3 \end{aligned}$$

Step 4. Solve.

$$\displaystyle \frac{2x^2 - 4}{-3} = 1 \;\Rightarrow\; 2x^2 - 4 = -3 \;\Rightarrow\; x^2 = \frac{1}{2} \;\Rightarrow\; x = \pm\frac{1}{\sqrt{2}}$$

Step 5. Check the con­di­tion $pq < 1$. $\displaystyle pq = \frac{x^2 - 1}{x^2 - 4} = \frac{-1/2}{-7/2} = \frac{1}{7} < 1$ for both roots, so both are valid.

Check­ing the answer#

For $\displaystyle x = \tfrac{1}{\sqrt{2}} \approx 0.707$: $\tan^{-1}(0.225) + \tan^{-1}(0.628) \approx 0.221 + 0.561 = 0.782 \approx 0.785$.

Answer#

$\displaystyle x = \pm\frac{1}{\sqrt{2}}$

Ques­tion 7: An equa­tion in inverse sines#

The prob­lem#

Solve $\displaystyle \sin^{-1}(1 - x) - 2\sin^{-1}x = \frac{\pi}{2}$.

Under­stand­ing the prob­lem#

Iso­late one inverse sine, take $\sin$ of both sides, and solve. Squar­ing-type steps can cre­ate false roots, so each root must be checked in the orig­i­nal equa­tion.

The idea#

Write $\displaystyle \sin^{-1}(1 - x) = \frac{\pi}{2} + 2\sin^{-1}x$, then use $\displaystyle \sin\left(\tfrac{\pi}{2} + \alpha\right) = \cos\alpha$ and $\cos 2\alpha = 1 - 2\sin^2\alpha$.

Step-by-step solu­tion#

Step 1. Rearrange and take $\sin$ of both sides. Let $\alpha = \sin^{-1}x$, so $\sin\alpha = x$.

$$\displaystyle 1 - x = \sin\left(\frac{\pi}{2} + 2\alpha\right) = \cos 2\alpha = 1 - 2\sin^2\alpha = 1 - 2x^2$$

Step 2. Solve.

$$\displaystyle 1 - x = 1 - 2x^2 \;\Rightarrow\; 2x^2 - x = 0 \;\Rightarrow\; x(2x - 1) = 0 \;\Rightarrow\; x = 0 \text{ or } x = \frac{1}{2}$$

Step 3. Check $x = 0$: $\displaystyle \sin^{-1}1 - 0 = \frac{\pi}{2}$. It works.

Step 4. Check $\displaystyle x = \tfrac{1}{2}$: $\displaystyle \sin^{-1}\frac{1}{2} - 2\sin^{-1}\frac{1}{2} = \frac{\pi}{6} - \frac{\pi}{3} = -\frac{\pi}{6} \ne \frac{\pi}{2}$. It fails.

Check­ing the answer#

The false root appeared because tak­ing $\sin$ of both sides loses infor­ma­tion: dif­fer­ent angles can have the same sine.

Answer#

$x = 0$

Com­mon mis­take to avoid#

Giv­ing both $0$ and $\displaystyle \tfrac{1}{2}$. Always sub­sti­tute back.

Ques­tion 8: $\displaystyle 2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}$#

The prob­lem#

Prove: $\displaystyle 2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}$.

Under­stand­ing the prob­lem#

Change $\sin^{-1}$ into $\tan^{-1}$ first, then use the dou­ble-angle for­mula for $\tan^{-1}$.

The idea#

$\displaystyle \sin\theta = \tfrac{3}{5}$ with $\theta$ acute gives a $3$-$4$-$5$ tri­an­gle, so $\displaystyle \tan\theta = \tfrac{3}{4}$. Then $\displaystyle 2\tan^{-1}x = \tan^{-1}\frac{2x}{1 - x^2}$ for $\lvert x \rvert < 1$.

Step-by-step solu­tion#

Step 1. Let $\displaystyle \theta = \sin^{-1}\tfrac{3}{5}$. Then $\displaystyle \cos\theta = \tfrac{4}{5}$ and $\displaystyle \tan\theta = \tfrac{3}{4}$, so $\displaystyle \sin^{-1}\tfrac{3}{5} = \tan^{-1}\tfrac{3}{4}$.

Step 2. Dou­ble it. $\displaystyle \tfrac{3}{4} < 1$, so the for­mula applies.

$$\displaystyle 2\tan^{-1}\frac{3}{4} = \tan^{-1}\frac{2 \cdot \frac{3}{4}}{1 - \frac{9}{16}} = \tan^{-1}\frac{\frac{3}{2}}{\frac{7}{16}} = \tan^{-1}\left(\frac{3}{2}\cdot\frac{16}{7}\right) = \tan^{-1}\frac{24}{7}$$

Check­ing the answer#

$\sin^{-1}0.6 \approx 36.87^\circ$; dou­bled, $73.74^\circ$; $\displaystyle \tan 73.74^\circ \approx 3.43 \approx \tfrac{24}{7}$.

Answer#

$\displaystyle 2\sin^{-1}\frac{3}{5} = 2\tan^{-1}\frac{3}{4} = \tan^{-1}\frac{24}{7}$.

Ques­tion 9: A sum of inverse cosines#

The prob­lem#

Show that $\displaystyle \cos^{-1}\frac{4}{5} + \cos^{-1}\frac{12}{13} = \cos^{-1}\frac{33}{65}$.

Under­stand­ing the prob­lem#

Let $\displaystyle A = \cos^{-1}\tfrac{4}{5}$ and $\displaystyle B = \cos^{-1}\tfrac{12}{13}$. We must show $\displaystyle \cos(A + B) = \tfrac{33}{65}$ and that $A + B$ lies in $[0, \pi]$.

The idea#

Use $\cos(A + B) = \cos A\cos B - \sin A\sin B$.

Step-by-step solu­tion#

Step 1. Both angles are acute. From the $3$-$4$-$5$ and $5$-$12$-$13$ tri­an­gles, $\displaystyle \sin A = \tfrac{3}{5}$ and $\displaystyle \sin B = \tfrac{5}{13}$.

Step 2. Com­pute.

$$\displaystyle \cos(A + B) = \frac{4}{5}\cdot\frac{12}{13} - \frac{3}{5}\cdot\frac{5}{13} = \frac{48}{65} - \frac{15}{65} = \frac{33}{65}$$

Step 3. $A$ and $B$ are acute, so $0 < A + B < \pi$, the range of $\cos^{-1}$. There­fore $\displaystyle A + B = \cos^{-1}\tfrac{33}{65}$.

Check­ing the answer#

$36.87^\circ + 22.62^\circ = 59.49^\circ$ and $\displaystyle \cos 59.49^\circ \approx 0.508 \approx \tfrac{33}{65}$.

Answer#

$\displaystyle \cos(A + B) = \frac{33}{65}$ with $A + B \in (0, \pi)$, so the iden­tity holds.

Ques­tion 10: Using the com­ple­ment prop­erty#

The prob­lem#

If $\displaystyle \sin^{-1}x + \sin^{-1}y = \frac{2\pi}{3}$, find $\cos^{-1}x + \cos^{-1}y$.

Under­stand­ing the prob­lem#

We are not asked for $x$ and $y$; we only need the sum of the inverse cosines.

The idea#

Use $\displaystyle \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$, so $\displaystyle \cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x$ (and the same for $y$).

Step-by-step solu­tion#

Step 1. Write each inverse cosine in terms of the inverse sine.

$$\displaystyle \cos^{-1}x + \cos^{-1}y = \left(\frac{\pi}{2} - \sin^{-1}x\right) + \left(\frac{\pi}{2} - \sin^{-1}y\right)$$

Step 2. Group.

$$\displaystyle = \pi - \left(\sin^{-1}x + \sin^{-1}y\right) = \pi - \frac{2\pi}{3} = \frac{\pi}{3}$$

Check­ing the answer#

Exam­ple: $\displaystyle x = y = \tfrac{\sqrt{3}}{2}$ gives $\sin^{-1}$ sum $\displaystyle \tfrac{\pi}{3} + \tfrac{\pi}{3} = \tfrac{2\pi}{3}$, and $\cos^{-1}$ sum $\displaystyle \tfrac{\pi}{6} + \tfrac{\pi}{6} = \tfrac{\pi}{3}$.

Answer#

$\displaystyle \frac{\pi}{3}$